Now to prove it he begins by taking a sequence $0 \rightarrow B \rightarrow B' \rightarrow B'' \rightarrow 0$ with $B'$ induced. Now since $H^2(G,B')=0$, there is a 1-cochain $f': G \rightarrow B$ such that $$u(x,y)=xf'(y)-f'(xy) + f(x), \qquad x,y \in G.$$
Then composing $f'$ with $B' \rightarrow B''$ we get a 1-cocycle $f'':G \rightarrow B$, with $d(\bar{f''})=\bar{u}$.
From which he deduces that $$\bar{s} \cup \bar{u} = \bar{s} \cup d(\bar{f''})=d(\bar{s} \cup \bar{f''})$$

Now by the previous lemma we know $\bar{s} \cup \bar{f''}= \bar{f''(s)}_0$ (here the subscript 0 denotes its class in $H^{-1}(G,B''))$. Now this is where I'm stuck he then says that $d(\bar{s} \cup \bar{f''})=d( \overline{f''(s)}_0)= \overline{N(f'(s))}^0$ (where the superscript 0 denotes its class in $\widehat{H}^0$), my question is why does the $f''$ change to $f'$? I cant see why this is so, he also does something similar in lemma 2 of this same appendix, but this is the lemma I need to use.

where $B_G = B/C$ with $C := \langle (g-1)b\mid g\in G,b\in B\rangle$ are the coinvariants and
$\delta=N$ is multiplication with the norm of $G$ (this is shown at the beginning of VI,§ 4 in Brown's book).